This is from Robinson, A Course in the Theory of Groups, 2nd edition:
Some pages below he gives an example:
In saying that $G=\langle x,y\mid x^2=1,y^2=1\rangle$, he is using the definition of presentation given in (2) in the picture. But a presentation, according to the author, is an epimorphism $\pi$ from some free group $F$ to $G$. In this example I want to find $\pi$. So, first I have to find the $Y$ and $S$ in definition (1), where $X=Y^{\pi}$ and then $F$ will be the free group on $Y$. Humm… $F$ I can already find it. It will simply be the free group on a set with two elements. But how do I find $Y$?
Best Answer
The free group $F=F(Y)$ on a set $Y$ has the defining property that given a function of sets $f\colon Y \to G$, where $G$ is (the underlying set of) a group, there is a homomorphism $\hat f\colon F \to G$ that extends $f$ in the sense that if $\iota\colon Y \to F$ is the natural inclusion (of sets), then $\hat f\iota = f$. Therefore to define the map $\pi$, it suffices to choose the image of the elements of $Y$ in $G$.
So let $Y = \{a,b\}$ be a two-element set and $G$ the infinite dihedral group as defined by the presentation $\langle x,y \mid x^2 = 1, y^2 = 1\rangle$. I will choose the function $Y \to G$ sending $a$ to $x$ and $b$ to $y$. The resulting homomorphism $\pi \colon F\to G$ is surjective because $\{x,y\}$ is a set of generators for $G$.